Trigonometry> what is the value of sin(37)cos(53)????an...

what is the value of sin(37)cos(53)????angles in degree

Hansraj Gyanendra Singh Rajawat , 9 Years ago

Grade 12

3 Answers

PASUPULETI GURU MAHESH

Last Activity: 9 Years ago

sin(90-37))The value of sin(37)cos(53) = sin(37)sin(90-37) (since cos(53)= sin(37)sin(37) =

= sin²(37) (since a.a=a²)

we know that sin (37)= 0.60 then sin²(37)= 0.36

Therefore the value of the sin(37)cos(53)=sin²(37)=0.36

Hansraj Gyanendra Singh Rajawat

Last Activity: 9 Years ago

I want to know how the value of sin 37 came???? this is my actual question how you calculate this value???????????

Ajay

Last Activity: 9 Years ago

Here is the solution..................................................................................................................

$Let\quad us\quad first\quad find\quad value\quad of\quad sin15\quad as\quad this\quad will\quad be\quad used\quad later.\\ we\quad know\quad cos30\quad =\quad 1-2{ sin }^{ 2 }15\\ This\quad will\quad give\quad us\quad sin15\quad =\quad \frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } \\ Now\quad use\quad the\quad formula\quad to\quad find\quad value\quad of\quad cos53sin37\quad as\quad follows\\ cos\alpha sin\beta \quad =\quad 1/2\left[ sin(\alpha +\beta )\quad -sin(\alpha -\beta ) \right] \\ Hence\quad cos53sin37\quad =\quad 1/2\left[ sin\quad 90\quad -sin\quad 15 \right] \\ Put\quad values\quad of\quad sin\quad 90\quad =\quad 1\quad and\quad sin15=\quad \frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } \\ cos53sin37\quad =\quad 1/2\left( 1-\frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } \right)$